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At rest, a rocket has an overall length of \(L .\) A garage at rest (built for the rocket by the lowest bidder) is only \(L / 2\) in length. Luckily, the garage has both a front door and a back door, so that when the rocket flies at a speed of \(v=0.866 c\), the rocket fits entirely into the garage. However, according to the rocket pilot, the rocket has length \(L\) and the garage has length \(L / 4\). How does the rocket pilot observe that the rocket does not fit into the garage?

Short Answer

Expert verified
Explain your answer. Answer: No, the rocket does not fit into the garage from the rocket pilot's perspective. This is because, from the pilot's point of view, the garage is contracted to a length of L/4 due to length contraction in special relativity, while the rocket is still L in length. Since the contracted length of the garage is shorter than the length of the rocket, the rocket cannot fit into the garage from the pilot's perspective.

Step by step solution

01

Calculate the Lorentz factor

First, we need to find the Lorentz factor, which is given by the formula \[ \gamma = \dfrac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \] Plug in the given value of \(v = 0.866c\), we obtain, \[ \gamma = \dfrac{1}{\sqrt{1 - \frac{ (0.866c)^2}{c^2}}} = 2\]
02

Calculate the contracted length of the rocket from the rocket pilot's perspective

According to the problem, the rocket pilot observes the rocket's length to be \(L\), which means there is no length contraction of the rocket from the pilot's perspective.
03

Calculate the contracted length of the garage from the pilot's perspective

Length contraction formula is given by \[ L' = \dfrac{L}{\gamma} \] where \(L'\) is the contracted length, \(L\) is the original length, and \(\gamma\) is the Lorentz factor. As the garage has a length of \(L/2\) when it is at rest, the rocket pilot observes the garage to have a length of \[ L' = \dfrac{L/2}{\gamma} = \dfrac{L/2}{2} = \dfrac{L}{4}\]
04

Determine whether the rocket fits into the garage

From the rocket pilot's perspective, the garage has a length of \(L/4\), while the rocket has a length of \(L\). Thus, the rocket does not fit into the garage from the pilot's point of view.
05

Conclusion

In this exercise, we have used the concept of length contraction in special relativity to examine the situation where the rocket pilot perceives their rocket to not fit into the garage. By calculating the Lorentz factor and the contracted lengths of the rocket and garage, we've shown that from the rocket pilot's perspective, the rocket doesn't fit into the garage because the garage appears to be only \(L/4\) while the rocket is still \(L\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lorentz factor
In the realm of special relativity, the Lorentz factor is crucial for understanding how time and space behave at high speeds. It's defined by the formula:
  • \( \gamma = \dfrac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
This factor comes into play when objects travel at velocities close to the speed of light \((c)\), causing their time, length, and mass to transform relative to an observer at rest.
In our rocket problem, the speed \(v = 0.866c\) results in a Lorentz factor of 2. This means that from the perspective of an outside observer, time appears to slow down, and lengths contract in the direction of the motion. Understanding this factor helps us calculate how these quantities change, revealing the hidden dynamics of objects moving at relativistic speeds.
Length contraction
Length contraction is a fascinating consequence of special relativity. When an object moves at a significant fraction of the speed of light, its length in the direction of travel seems shorter to a stationary observer. This is given by the formula:
  • \( L' = \dfrac{L}{\gamma} \)
Here, \(L'\) is the contracted length, \(L\) is the original length, and \(\gamma\) is the Lorentz factor.

In the exercise, the rocket's pilot sees no contraction of the rocket because they are moving with it, so its length remains \(L\). However, the garage appears shorter to the pilot due to its relative motion. Calculating with the Lorentz factor of 2, the garage's length contracts to \(L/4\) from the pilot's viewpoint.

This concept highlights how observers in different frames perceive different realities, each equally valid in their own context.
Rocket paradox
The rocket paradox elegantly demonstrates relativity's counterintuitive nature. From the stationary observer's frame, the rocket fits into the garage due to length contraction, giving the illusion of it being shorter.
However, from the rocket pilot's perspective, the garage is the object in motion, and thus it experiences length contraction, appearing only \(L/4\) long.

This paradox illustrates the core tenets of special relativity, where an observer in motion perceives the dimensions of stationary objects differently. It showcases how reality can differ substantively for observers in different frames, challenging our everyday intuitions about space and movement.

By exploring this paradox, we gain insight into the flexible and interconnected nature of time and space, where both perspectives coexist without contradiction in the relativistic world.

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Most popular questions from this chapter

In Jules Verne's classic Around the World in Eighty Days, Phileas Fogg travels around the world in, according to his calculation, 81 days. Due to crossing the International Date Line he actually made it only 80 days. How fast would he have to go in order to have time dilation make 80 days to seem like \(81 ?\) (Of course, at this speed, it would take a lot less than even 1 day to get around the world \(\ldots . .)\)

Using relativistic expressions, compare the momentum of two electrons, one moving at \(2.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\) and the other moving at \(2.00 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\). What is the percentage difference between classical momentum values and these values?

Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means \(d v / d t=g,\) where \(g\) is a constant, since that would have \(v\) exceeding \(c\) in a finite time. Rather, it means that the acceleration experienced by the moving body is constant: In each increment of the body's own proper time \(d \tau,\) the body acquires velocity increment \(d v=g d \tau\) as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation: a) Write a differential equation for the velocity \(v\) of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation, remembering that squares and higher powers of differentials can be neglected. b) Solve this equation for \(v(t),\) where both \(v\) and \(t\) are measured in the ground frame. c) Verify that your solution behaves appropriately for small and large values of \(t\). d) Calculate the position of the body \(x(t),\) as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time \(t=0,\) at ground-frame position \(x=c^{2} / g\) e) Identify the trajectory of the body on a space-time diagram (Minkowski diagram, for Hermann Minkowski) with coordinates \(x\) and \(c t,\) as measured in the ground frame. f) For \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) calculate how much time it takes the body to accelerate from rest to \(70.7 \%\) of \(c,\) measured in the ground frame, and how much ground-frame distance the body covers in this time.

Use the relativistic velocity addition to reconfirm that the speed of light with respect to any inertial reference frame is \(c\). Assume one-dimensional motion along a common \(x\) -axis.

Use light cones and world lines to help solve the following problem. Eddie and Martin are throwing water balloons very fast at a target. At \(t=-13 \mu s,\) the target is at \(x=0,\) Eddie is at \(x=-2 \mathrm{~km},\) and Martin is at \(x=5 \mathrm{~km},\) and all three remain in these positions for all time. The target is hit at \(t=0 .\) Who made the successful shot? Prove this using the light cone for the target. When the target is hit, it sends out a radio signal. When does Martin know the target has been hit? When does Eddie know the target has been hit? Use the world lines to show this. Before starting to draw your diagrams, consider: If your \(x\) position is measured in \(\mathrm{km}\) and you are plotting \(t\) versus \(x / c,\) what units must \(t\) be in, to the first significant figure?

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